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PORun one Jquery Script independently for each <div>
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<p>I have the following question:</p> <p>I am working on a JQuery script that turns some HTML into a "Slider". No problem so far, everything's working smoothly.</p> <p>The thing is is, if I want to run two or more "sliders" at the same time, of course they all "act" the same.</p> <p>If I want them to act differently at the same time, can I "add" the script several times in the HTML-Code (, or do I have to add a ".each()" function, to the script?</p> <p>HTML-Code:</p> <pre><code><div id="galleria1" interval="1000"> <a href="#" class="show" > <img width="640" height="450" rel="Slider-Image"src="a.jpg" /> </a> ... </div> <div id="galleria2" interval="1000"> <a href="#" class="show" > <img width="640" height="450" rel="Slider-Image"src="b.jpg" /> </a> ... </div> </code></pre> <p>Would be cool of any of you knows a solution to this!</p> <p>Thanks, Mario.</p> <p>EDIT:</p> <p>Here the Script:</p> <pre><code>$(document).ready(function() { //checks if page is completely loaded slideshow(); }); function slideshow() { //Sets all pictures opacity to 0 $('#galleria a').css({opacity:0.0}); //Sets the first picture's opacity to 1 $('#galleria a:first').css({opacity:1.0}); //Calls the gallery() function to run the slideshow, number gives time till next 'slide' im milliseconds setInterval('nextPicture()', 1000); } function nextPicture() { //If no image hast the class 'show', takes the first picture var current = ($('#galleria a.show')? $('#galleria a.show') : $('#galleria a:first')); //Get next image, if reached the end of slideshow, starts from the begining var next = ((current.next().length == 0) ? $('#galleria a:first') : current.next()); //Runs Animation and makes next picture visible, number gives fade in time next.css({opacity: 0.0}) .addClass('show') .animate({opacity: 1.0}, 1000); //Hides last picture current.animate({opacity: 0.0},1000) .removeClass('show'); } </code></pre> <p>-> used the same id="galleria" for both so far. (already changed this in the html above)</p> <p>Thanks for the answers so far!</p>
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<javascript><jquery><html><css>
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Run one Jquery Script independently for each <div>
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